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Construction: (construction Of A Triangle Equal In Area Of A Quadrilateral)

Class 9th Mathematics West Bengal Board Solution
Draw And See 14
  1. Pritam drew a quadrilateral ABCD of which AB = 5 cm, BC = 6 cm, CD = 4 cm, DA = 3 cm…
  2. Sahana drew a quadrilateral ABCD of which AB = 4 cm, BC = 5 cm, CD = 4.8 cm, DA = 4.2…
  3. Sahana drew a rectangle ABCD of which AB = 4 cm and BC = 6 cm. I draw a triangle with…
  4. I draw a quadrilateral ABCD of which BC = 6 cm, AB = 4 cm, CD = 3 cm, ∠ABC = 60°, ∠BCD…
  5. I draw a square with side 5 cm. I draw a parallelogram with equal area of square of…
  6. I draw a square with side 6 cm. and I draw a triangle with equal area of that square.…
  7. I draw a quadrilateral ABCD of which AD and BC are perpendicular on side AB and AB = 5…
  8. I draw any pentagon ABCDE and draw a triangle with equal area of it of which one vertex…
Draw And See 14
Question 1.

Pritam drew a quadrilateral ABCD of which AB = 5 cm, BC = 6 cm, CD = 4 cm, DA = 3 cm and ∠ABC = 60°, I draw a triangle with equal area of that quadrilateral.
Answer:

A. Draw the given quadrilateral ABCD.



B. Draw the diagonal DB of quadrilateral ABCD.


C. Draw a parallel line through point A to diagonal DB of quadrilateral ABCD which intersects at F produced BC.


D. Join FD, AF||BD


ΔDFC is the required triangle.



Proof:


∆ABD = ∆BFD (on same base DB and between same parallels DB and AF)


∴ ∆ABD = ∆BFD


∆DBC + ∆ABD = ∆BFD + ∆DBC (adding area of ∆DBC on both sides)


∴ quadrilateral ABCD = ∆DFC
Question 2.

Sahana drew a quadrilateral ABCD of which AB = 4 cm, BC = 5 cm, CD = 4.8 cm, DA = 4.2 cm and diagonal AC = 6 cm. I draw a triangle with equal area of that quadrilateral.
Answer:

A. Draw the given quadrilateral ABCD.


B. Draw the diagonal AC of quadrilateral ABCD.



C. Draw a parallel line through point D to diagonal AC of quadrilateral ABCD which intersects at F produced BC.


D. DF||AC, ABF is the required triangle.



Proof:


∆ADC = ∆ACF (on same base AC and between same parallels AC and DF)


∴ ∆ ADC = ∆ ACF


∆ABC + ∆ADC = ∆ACF + ∆ABC (adding area of ∆ABC on both sides)


∴ quadrilateral ABCD = ∆ABF
Question 3.

Sahana drew a rectangle ABCD of which AB = 4 cm and BC = 6 cm. I draw a triangle with equal area of that rectangle.
Answer:

A. Draw the given rectangle ABCD.



B. Draw the diagonal AC of rectangle ABCD.


C. Draw a parallel line through point D to diagonal AC of rectangle ABCD which intersects at F produced BC.


D. DF||AC, ΔABE is the required triangle.



Proof:


∆ACE = ∆ADC (on same base AC and between same parallels AC and DE)


∴ ∆ACE = ∆ADC


∆ABC + ∆ADC = ∆ACE + ∆ABC (adding area of ∆ABC on both sides)


∴ rectangle ABCD = ∆ABE
Question 4.

I draw a quadrilateral ABCD of which BC = 6 cm, AB = 4 cm, CD = 3 cm, ∠ABC = 60°, ∠BCD = 55°, I draw a triangle with equal area of that quadrilateral of which one side is along side AB and other side is along side BC.
Answer:

A. Draw the given quadrilateral ABCD.



B. Draw the diagonal AC of quadrilateral ABCD.


C. Draw a parallel line through point D to diagonal AC of quadrilateral ABCD which intersects at E produced BC.


D. DE||AC, ΔABE is the required triangle.



Proof:


∆ACE = ∆ADC (on same base AC and between same parallels AC and DE)


∴ ∆ACE = ∆ADC


∆ABC + ∆ADC = ∆ACE + ∆ABC (adding area of ∆ABC on both sides)


∴ quadrilateral ABCD = ∆ABE
Question 5.

I draw a square with side 5 cm. I draw a parallelogram with equal area of square of which one angle is 60°.
Answer:

A. Draw the given square ABCD.



B. Draw the diagonal DB of square ABCD.


C. Draw a parallel line through point C to diagonal DB of square ABCD which intersects at E produced AB.


D. CE||DB.



Proof:


Since sides of square are equal and parallel, AB||DC and thus, DC||AE.


∴ square ABCD = ∆ABE (on same base DC and between same parallels DC and AE)
Question 6.

I draw a square with side 6 cm. and I draw a triangle with equal area of that square.
Answer:

A. Draw the given square ABCD.



B. Draw the diagonal DB of square ABCD.


C. Draw a parallel line through point C to diagonal DB of square ABCD which intersects at E produced AB.


D. CE||DB. ΔADE is the required triangle.



Proof:


∆DCB = ∆DBE (on same base DB and between same parallels DB and CE)


∴ ∆DCB = ∆DBE


∆ABD + ∆DCB = ∆DBE + ∆ABD (adding area of ∆ABD on both sides)


∴ square ABCD = ∆ADE
Question 7.

I draw a quadrilateral ABCD of which AD and BC are perpendicular on side AB and AB = 5 cm, AD = 7 cm and BC = 4 cm. I draw a triangle with equal area of that quadrilateral of which one angle is 30°.
Answer:

A. Draw the given quadrilateral ABCD.



B. Draw the diagonal DB of quadrilateral ABCD.


C. Draw a parallel line through point C to diagonal DB of quadrilateral ABCD which intersects at Q produced AD.


D. CQ||DB, ΔABQ is the required triangle.



Proof:


∆DCB = ∆DBQ (on same base DB and between same parallels DB and CQ)


∴ ∆DCB = ∆DBQ


∆ABD + ∆DCB = ∆DBQ + ∆ABD (adding area of ∆ABD on both sides)


∴ quadrilateral ABCD = ∆ABQ


Taking BQ as a base, we draw another ∆BCE with one angle 30⁰ and between the parallels DQ and BC.


∴ ∆BCE = quadrilateral ABCD
Question 8.

I draw any pentagon ABCDE and draw a triangle with equal area of it of which one vertex is C.
Answer:

A. Draw the given pentagon ABCDE.



B. Draw the diagonal AC of pentagon ABCDE.


C. Draw a parallel line through point E to diagonal AC of pentagon ABCDE which intersects at F produced AB.


D. AC||FD, ΔABC is the required triangle.



Proof:


∆ACF = quadrilateral ACED (on same base AC and between same parallels AC and FD)


∴ ∆ACF = quadrilateral ACED


∆ABC + ∆ACF = quadrilateral ACED + ∆ABC (adding area of ∆ABC on both sides)


∴ ∆ABC = pentagon ABCDE